Another Approach to Titration Curves Which is the Dependent Variable? Christopher J. Willis University of Western Ontario. London. Ontario N6A 587, Canada
T h e treatment of acid-hase equilibrium in aqueous solution has long been an important part of the general chemistry curriculum. Many instructors, however, are continually frustrated a t the level of competence achieved by students in solving numerical problems, even a k e r considerable exposure to the concent.s. In the author's exnerience. the difficultv faced by the s t u d ; n t i s not so much mathematical a s it is'in the recognition o f t h e approach required in different situations. Do we have a solution ofastrong or weak acid, a strong or weak base, or a hull'er mixture? Are we a t the equivalence point, or is one or the other reactant in excess? Can t h e degree of dissociation be considered nerlirible? Is hvdrolvsis . . auureciahle'? I h e s the dissociation of water contribute signific&ly t o the ion roncentrations? These questions, with the value judgments implicit in such terms as "negligible," "appreciahle," or "significant," can bewilder a student k i n g anything hut a routine situation where a rote-learned formula may be a p plied. T h e problems are particularly acute in t h e calculation of a titration curve. Taking the familiar example of a strong hase added to a weak acid, i m r different regions must be identified; the initial acid solution ("weak acid dissociatim"), the region I~eforethe eauivalenre m i n t . where the acid is oartiallv neut.ralized i"h&'er solutihn"), the equivalence p k n t i"h;drovsis"). and the region following the equivalence point ("strong base"). T h e student has t o recognize each of these situations and L o decide on the appropriate method of attack including apprtrximations. I~eforeheginning any numerical calculation. The hetter student in particular will he worried ahout the way in which these four apparently independent calculation methods can fit together to make a smooth curve. Surely, the student asks, "Is there some general equation which could be used t o generate the whole curve from one consistent set of assumotions?". T h e answer we usuallv rive t o t h e student is t h a t a n exact treatment can he applied to acid-hase equilibria, resulting in equations applicable to anv situation. However, their numerical solution is too rom~plicatedfor everyday use, since finding p H from known values of'molarities and dissociation constants usually involves solving a high-order polynomial. Therefore, simplifying assumptions are used, hearing in mind t h a t different terms in the oolvnrmials will he nerlieihle in . . . .. different situations. T h e various equations are available in the literature, having been discussed in THIS dOLIRNA1. ( 1 ). in the average student in'this proillem. ~ r e n e m a n(4j h a s d e scrihed a n iterative routine hy which a titration curve may he plotted for pdyprutic acid having successive K , values, h u t major computer facilities are required. Various authors ( 5 ) have suggested simpler programs, often suitahle for a programmahle pocket calculator, by which specific problems in acid-base equilihrium may be solved. In this paper, a n approach to titration curves is descrihed which eliminates two of' the stumbling hlocks mentioned Presented in part at the 2nd Chemical Congress of the North American Continent. Las Vegas. NV. August 1980.
above; the choice of approximations and the algebraic cornplexitv of accurately solving for [H+].T h e technique is very simple. Instead of attempting t o calculate pH in a solution made up in specified manner, we regard moles of titront as the dependent variahle and ralrulate titrant added ns a functian r ~ f p H In . other words, the question asked is: "How much reactant has t o he added in order to reach this pH?" rather than: "What is the p H o f t h e solution after such-andsuch volume of reactant has heen added?" A moment's thought shows t h a t the first. method ofasking the question is perfectly logical and consistent with experimental practice. Indeed, in all acid-hase titrations performed with a n indicator (as opposed t o a pH meter), i t is prr~cisel? this quantity which is mrnsurcd a n d r r r o r d ~ d - t h e volume oftitrant required t o reach a p H whosevalue has been previw s l v snecified hv the choice of indicator. 1" orher t o show why t h e calculation of a titration curve is simplified by this method of approach, let us consider the familiar case o f s t r m g hase ROH added to weak acid HA. (The calculation is more straightforward if it is assumed that the total volume of the solution remains constant throughout.) LJsing IH+I t o represent total hydronium ion concentration. we h& t h o s u i equilihria:
,...., Suppose M,, moles of HA were originally dissolved in each liter ofsolutim, then a mass halance gives:
+
M. = [HA\ [A-I i ?) If HOH is a strong hase, completely ionized into H+ and OH-, the charge balance in the solution after addition of HOH requires: [H+l + lH+l = [A-1
+ IOH-I
(3)
Eliminating lHAl hetween eqns. (1) and (2) gives: [A-1 =
K,M,, IH+I + Ka
(4)
Suhstituting in eqn. (3) and writing K,,,I[H+]for [OH-] (where K,,. is the ion product for water) gives: .
.
.
.
Several points about this equation may he noted: ( 1 ) Its derivation requires no approximation. It accurately describes the hehavior of the solution at all times durinp the tilration. ( 2 ) [Btl is derived from the number 01' moles of BOH added to I I of solutir,n and the ratio d [ H t I to M. provides a convenient indication of the progressive stoichiometry of mixing as the Litralim progresses. The equivalence point is reached when IB+] = M,. (31 in the usual "strong acid" calculation. HA is considered completely disswiated at all times. Fmm eqn. 14).this is equivalent tosaying K , >> [Htl, so that K , + [Ht] * K . and [A-1 = Mo. At the equivalence puint, [Btl = [A-1 = M., and. f n m eqn. (5). IH+12= K,,. (41 For H yiven value of [Btl ii.e. moles of strong hase added! the ct,rresponding value of lHtl may he found. hut acubic in [Ht[ must he wived. However. the r r w r s e process, calculatine [Bt] from I H t l is relatively trivial. Volume 5 8
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By the repeated use uf eqn. (51, therefore, a complete titration curve may he developed using computing facilities no more complicated than a irogrammable bock& calculator. Values of p H are inserted a t chosen intervals of, say, 0.2 p H unit, over the range of p H of interest, and corresponding [Rf] values are calculated using a simple routine and constant, stored, values uf K,, M,, and K,,,.If pH changes are followed over the range 1.0-13.0, a total of 66 points will be obtained, from which a n accurate titratiun curve may be plotted. Despite the manner of calculation, the points will still be plotted most conveniently in t h e usual manner with [R+] a s t h e abcissa and p H the ordinate, the only unusual feature being that in t h e vertical, rather than horthe points are evenlv . spaced . izontal, direction. If sliehtlv more sophisticated computing facilities are availahie, the more sen.ior student may he asked t o write and run a program t o perform a repeated calculation with incrernented values of p H , or t o plot a complete titration curve. Such exercises in t h e use of a simple algorithm t o generate a curve are hecuming increasingly important as we train students t o enter the world of the microcomputer and it9 capacity for data-processing. At the same time t h a t [B+]is being calculated from pH, i t is easy and instructive t o calculate [A-] and [HA] using eqns. (4) and 12). By plotting these two concentrations as a function of [B+],the neutralizatiun of the acid as the titratiun proceeds may be illustrated without difficulty. Figure 1shows the result of applying this calculation t o acetic acid, M, = 0.100, K, = 1.8 X 1 P . Several paints emerge in the course of this calculation:
lation of the initial pH, but it is worth noting that the titration curve may be extrapdated back through this point. ( 2 ) The pH shows the expected rapid change near the equivalence ooint. The result lHil = 0.100 isobtainedfromeun. ( 5 ) bv in^ &ting [Ht] cc,rre&nding to pH values in the range 72. to i0.6. The pH at the equivalence point may be estimated hy averaging these two values (or graphically from the titration curve) and the result. 8.9, agrees with that ohtained by the usual "hydrolysis" calculation cm U.lOO M sodium acetate solution (pH calculated, 8.87). The method of calculation outlined above may be extended t o polyprotic acids. In the case of a diprutic acid, H2A, having successive ionization constants K I a n d K2, t h e mass balance equation becumes: M, = [HsA]
+ [HA-] + [A2-]
(61
~
(11 Attempts to solve eqn. (5) for pH values helow 2.88 give negative values of [Bt]. The physical significance of the addition of a "negative" amount of base is, of course, that the corresponding amuunt of strong acid would have to he added to reach these low pH values. In other words, the actual titration starts s t the point where (Bi] = 0 (pH 2.881 which is, of course. the pH value of 0.100 M acetic acid. An astute student will at once see that some effort can he saved by a preliminary calcu-
Figure 2. The titration of phosphwous acid (pK, = with strong base.
Figure 1. The titration of acetic acid (pK. = base.
660
4.74). 0.100
Journal of Chemical Education
M with strong
1.30.pK2 = 6.59).0.100 M.
Figure 3, The titration of succinic acid (pK, = 4.21, pK, = 5.64). 0.100 M. with strong base.
from which it may he shown that:
M.KtKz (9) = [ H + ] ? + K 1 [ H t J + KIKp The charge balance eouation after addition of strong hase HOH is:
IA'?
[B+l + [H+I = [HA-]
+ 2[AZ-I + (OH-]
(101
and the suhstitution gives:
As hefore, we have an equation where the calculation of [B+] from [H+] is relatively straightforward, hut the calculation of [H+] from [B+] is very difficult. The application of this equation to t i ) representative diprotic acids is illustrated in Figures 2 and 3. Phosphorous acid, H:,PO:*(Fig. 2) has well-separated ionization constants (KI = 5 X 1W2,K? = 2.6 X Asaresult, two end points are seen in the titration curve and the plot of the concentrations of the various species present shows that virtually all HXPO:Ihas been converted into H ~ P O J -before any appreciable concentration of HPOa2- appears. In calculations on this system hy other methods, thissimplifies matters, since the successive dissociations may he treated independently. In succinic acid (Fig. 3), hy contrast, the two ionization constants are not well separated (Kt = 6.2 X lo-", K:, = 2.3
Figure 4. The titration of a mixture of formicacid IpK. = 3.74). 0.100 acetic acid (pK. = 4.74). 0.100 M. with strong bare.
M, and
,. ...
Y i n-sr,
As a result, there is considerable formation of the dianinn Succ" before all H:,Succ has been ionized to HSucc-, the respective percentages a t the half-neutralization point heing 14% 14%, and 52% Equilibrium hetween these three effectively buffers the solution in this region, and no end point can he detected. Since the two ionization processes can no longer be considered independently, the calculation of ion concentrations in this region hy conventirmal methods is too complex for the average student, hut the use of eqn. (11) enables the curve to he determined without difficulty. A related aoolication of this method is to the titration of a .. mixture 01 t w o cnr inore ardx 111sdution together. Suppose the a c d s a n H A and HI\' u.ith res~)~cti\~t~~~~~~~centrati~ns M,, and M,' and dissociation constan& K, and K,'. The char& balance equation after the addition of strong hase BOH will be IRf]
+ IHtl
= [A-I
+ [A'-] + [OH-]
112)
Substitution in terms of molarities and ionization constants ~ives:
which is, of course, analogous to eqn. (5).As with the diprotic acid, two situations arise. If the two ionization constants are well senarated. the two eauilihria mav he treated successivelv. the nekralization of the stronger acid heing completed before aovreciahle ionization of the weaker has heeun. If. however. .. the two ionization constants are relatively close, the neutralization process must he treated simultaneously, and the calculation of pH changes near the end point is complex. Use of eqn. (13) enahles (B+I changes necessary to reach each pH value to he calculated without difficulty. As an example, the titration curve of an equimolar mixture of formic acid IK, = 1.8 X lo-') and acetic acid (K, = 1.8 X 10-" is shown in Figure 4. As would he expected, with a difference of only a factor of 10 lone pK, unit) in K. values, no detectahle "first" end point, corresponding to the neutralization of the stronger acid, can he seen. The plot of ionic concentrations shows that hoth acids
Figure 5. The titration of s mixture of periodic acid (pK, = 1.64). 0.100 M. and hydrazoic acid (pK, = 4.72). 0.100 M. wilh strong base. are gradually ionized throughout the titration, and the end point is seen only when hoth have been neutralized. Even when the two acids differ appreciably in acidity by as much as three pK, units, a second end point is barely detectable. Figure 5 shows a curve calculated for equimolar periodic acid (KO = 2.3 X 10-2) and hydrazoic acid (K. = 1.9 X lo-". While the plot of ionic concentrations shows a reasonable decree of senaratirm between the successive neutralization of t he stronger :~ndweakrr acid, the titration curve shows that the increaw it1 ~ J dHl 1he first euuit~alenrerwint is t(m shitllow to be experimentally detected by the use bf an indicator, although it would he found hy following the course of the titration with a pH meter. As will he appreciated, calculation of equilibrium in a mixed solution of this type is generally considered so complex as to preclude its discussion with a class. By treating IH+] as the independent variahle in eqn. ( l a ) , the calculation of the titration curve is made conceptually easy and mechanically Volume 58 Number 8
August 1981
661
straightforward f i r the student, who may then see the relationship hptween changing concentrations of various species and pH as the titration proceeds. It will be seen easily that eqn. (13) may be extended to include any number of weak acids in the solution by the addition of a term of the type K,M,I(IH+] K, for each component. No discussion of titration curves would he complete without inclusion r ~ fa weak hase equilihrium. Since virtually all compounds of this class are proton acceptors (NH:Iand related species), the additional equilibrium may he represented as
+
A mass balance gives [HI + IHBtl = Mh
(15)
where Mh is the numher rd'mdes ofweak hase added to 1 1 of the salutinn of' acid in titration. Comhination of these two equations gives: , From charge halance. [HB+] takes the place of [R+] in the strong hase equilibria considered previously, so the progress of the titration can he followed hy calculating [HB+],and from it Mh, as a function of [Ht]. As illustration, consider aqueous ammonia. NH,(aq),added to a weak acid. From eqn. (51, we have the expression for [HB+] (which is, of course, [NHd+] in this case): K M I(f,, IHBtI = I H + ~ + +-~ , lHil and combining this with eqn. (16) gives
W+l
(171
This equation. which is independent of K,, represents the titration curve fur ammonia or other weak base added to strong acid, the latter being assumed completely dissociated. Figure 6 shows the titration curve calculated fhr NH:,iaq1 (Kh = 1.8 X 10Y') added to strong arid (K, = 10) using eqn. (18). A good end point is ohtained, the result Mh = 0.100 being obtained by inserting values of [H+]corresponding to pH from 3.4 to 6.8 in eqn. (18). The midpoint of this range, pH 6.1, corresoonds to the DH ohtained bv the usual "hvdrulvsis" " . calculation for an 0.100 M solution of an ammonium salt of a strong acid (calculated, pH 5.13). Rv contrast, the titration of aqueous ammonia against acetic acid (Fig. 7 ) shows ntr sharp end point. The calculation of this titration curve by conventional routes is seldom undertaken, since the simultaneous equilibria of weak acid and weak hase make it difficult to use aily simplifying assumption, particularly in the neighborhood of the equivalence point. Use ot'eqn. (18) enables the curve to he determined without difficulty and shows the solution to he neutral a t the equivalence point. This result would, of course, he intuitively expected from the accidental coincidence of the numerical values rrf K, for acetic acid and Kh for aqueous ammonia; the usual formula for calculating pH in a solution of a salt where acid and hase are hoth weak is [H+] = \/K,&IKh. As a final examnle of the aoolication of this method of
-
.
obtjined easily by the comhination of eqns. (11) and (161, giving
Note two possible simplifications of this equation. By assuming KhlH+I >> K,,,.the term (KhIH+I K,,)IKhlH+l . . .tends to unity, and eqn. (18) reverts t > [Hf],the term K,M,I([Hf] K,) tends ~ I M, I and eqn. (18) simplifies to:
The application of this equation to the titration of phosphorous acid with aqueous ammonia is shown in Figure 8. Comparison with Figure 2, in which the same acid was titrated with strong hase,shows that the first end point is very similar and would he readily detectable in practice. The second end point, however, is now nonexistent; whereas, with a strong hase it was well-defined. A titration curve of this type nicely brings out the point that a weak hase will give a detectable end print only with a fairly strong acid (phosphorous acid has pK1 = 1.30), not with a weak acid (pK2= fi.58). Various other types of titration curve may, of course, he
0.100 M. with the weak
Figure 7.The titration of acetic acid (pK, = 4.74).0.100 M. withaqueous ammonia (pK, = 4.74).
+
+
Figure 6.The titration of a strong acid IpK, = -1.0). base aqueous ammonia (pK, = 4.74).
662
Journal of Chemical Education
constructed hy using this approach with the appropriate equation. Equation (11) may he extended to a triprutic acid to give:
It is rarely worthwhile tu plot a titration curve fur such an acid, however, since the third ionization constant is generally so small that no end point is detectable (ex., H:$P04,pKa = 12.4). A complementary set of equations may, of course, he constructed for the reverse mode of titration, in which the acid is added tu the base. All that is necessary is the rearrangement of the appropriate equation given ahove so that M, (moles of acid added) isexpressed in termsof [B+Jor M h (initialmoles of hase), the appropriate K , and Kh values, and [H+]. In summary, the following advantages may he claimed for this method of calculating titration curves: (1) No appruximatiansare used: all equations are exact. NoUvalue judgements" are required. 121 Each curve is calculated cmtinoouslv usine one eauation. are all that are needed. (4) Evaluations of si~nultaneousequilibria involving components of similar acidity present no difficulty. (5) The concentrations of any or all uf the species present in solotion may he extracted easily h m the data as the calculation
Figure 8. The titration of phosphorous acid (pK, = with aqueous ammonia (pKb = 4.74).
1.30.pK2 = 6.59) 0100 M.
proceeds. 16) Using evaluated ameentrations, such statistics as "degree of ionization" or "extent of h.vdn$ysis" may he calculated at any point of interest in the titration. (7) The logic of the questim asked (how much reagent must he added to reach a specified pH?) corresponds exactly to the
student's laboratory experience. 181 Necessary computing facilities are readily available at the nockrt cnlculatw level: should the student have access to more saphislicat~dfacilities, the calculations lend themselves to a useful exercise in pmgrammin~.
.~ ~
~
(Programs for the solution of the equations described here require 50-100 steps and up to 9 memories on a pocket calculator. The author would be glad t o supply copies to anyone interested.) There will, of course, always he a place in the teaching of elementary solution chemistry for the use of "approximate" methods of calculation, and the traditional approaches will often he best for solving single numerical problems on simple systems. T h e intention of this article has been to suggest that the availability of the proprammahle pocket calculator gives us the upport&ty t o p s s e n t familiar material to students in new ways, either by preparing titration curves in advance
for discussion with the class, or, with a suitable group of students, assigning them the task. Bv remuvine both conceotual and mechanical harriers to the r ; ~ l w l : ~ t i ot'titration ot~ vurwi. s e ma\ :boa rt~ldt'ot.how t,, I I ~ I I : ~ I Iim I wrrall v i r w of i~ci(l-ha+ w u ~ l ~ l ~ r iiuu nmvarier\ of solutions. After they have come appreciate general trends, they will appreciate what approximations may he made in a given situation, a logical process more satisfactory than the usual piecemeal approach in which the approximations must come first. It is unlikely that we shall ever succeed in making acid-base equilibrium an easy topic for every student, but we should he continually trying to improve our presentation of this important material.
,
... . ,.
(2,Hutlrr, .I.
N.."lun8e Fquilihriurn. A Mathernnllcal Appnmh."Addiim~W~riry. Ra~diny.
M a l l . 1964.
M.. "Aud Heslren~thand P r t ~ h l v rl'urrci i~ ~n Rater". in "Treatirc m Analytirai Vhemirrrv" i6d8rrrr tiiithotf. I. M Eirmne. 1'. .I..and Sandell. E.B i Inrrrrrienre. New Y w k . N.Y.. 1959. Chxplrr 12 l i i Hrenernan.1:. L,,.J.CHaME n r ' C . . S l . X l 2 l i 9 i l i l i i See, r . ~ .Willialna. . H. 1'..1. CHliM. E1111C..66.?:37 119791. l.ii see. O X . , Hruckcnr~in.S.md Kolthoff. 1.
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August 1981
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